EXPLANATION OF GOLD IONIZATIONS
By Prof. L. Kaliambos (Natural Philosopher in New Energy) June 21 , 2015 ' INTRODUCTION' Gold is a chemical element with symbol Au and atomic number 79. However despite the enormous success of the Bohr model and the quantum mechanics of Schrodinger in explaining the principal features of the hydrogen spectrum and of other one-electron atomic systems, so far neither was able to provide a satisfactory explanation of ionizations of many electon atoms related to the chemical properties of atoms. Though such properties were modified by the periodic table initially proposed by the Russian chemist Mendeleev the reason of this subject of ionizations of elements remained obscure under the influence of the invalid theory of special relativity. (EXPERIMENTS REJECT RELATIVITY). It is of interest to note that the discovery of the electron spin by Uhlenbeck and Goudsmit (1925) showed that the peripheral velocity of a spinning electron is greater than the speed of light, which is responsible for understanding the electromagnetic interaction of two electrons of opposite spin. So it was my paper “Spin-spin interactions of electrons and also of nucleons create atomic molecular and nuclear structures” (2008), which supplied the clue that resolved this puzzle. Under this condition we may use this correct image of Gold including the following ground state electron configuration: 1s22s22p63s23p63d104s24p64d105s25p6 4f145d106s1 According to the “Ionization energies of the elements-WIKIPEDIA” the ionization energies (eV) of gold (from (E1 to E3 ) are the following: E1 = 9.22 , E2 = 20.5 , and E3 = 30 . For understanding better such ionization energies see also my papers about the explanation of ionization energies of elements in my FUNDAMENTAL PHYSICS CONCEPTS. Moreover in “User Kaliambos” you can see my paper of 2008. EXPLANATION OF E1 = 9.22 eV = - E(6s1) According to the experiments the electrons of 6s1, 5d10 and 4f14 belong to the same energy level under n = 6. However the electron charges (-77e) of the 77 electrons of the following configuration (1s22s22p63s23p63d104s24p64d105s25p64f 145d10) screen the nuclear charge (+78e) and for a perfect screening the electron of 6s1 would provide an effective ζ = 1. Of course the one electron ( 6s1) penetrates the 5d9 and leads to the deformation of clouds. Thus ζ > 1. Here the E(6s1) represents the binding energy of 6s1given by applying the Bohr formula as E1 = 9.22 eV = -E(6s1) = -(-13.6057)ζ2/n2 Then using n = 6 the above equation can be written as E1 = 9.22 eV = -E(6s1) = -(-13.6057)ζ2/ 62 and solving for ζ we get ζ = 4.94 > 1 . ' ' EXPLANATION OF E2 = 20.5 eV = -E(5d2) + E(5d1) It is of interest to note that the 5d10 consists of five pairs (10 electrons of 5d2, 5d2, 5d2, 5d2 , and 5d2 with opposite spin). Here the E(5d2) represents the binding energy of the first 5d2, while the E(5d1) represents the binding energy of 5d1, which appears after the first ionization of the first 5d2 . Here the charges (-68e) of the following electron configuration (1s22s22p63s23p63d104s24p64d105s25p64f14) screen the nuclear charge (+79e ) and for a perfect screening we would have an effective ζ = 11. However the electrons of 5d10 penetrate strongly the 4f14 and provide an effective ζ < 11. Note that the first 5d2 consists of one pair (2 electrons of opposite spin). Thus applying my formula of 2008 we write -E(5d2) = -+ (16.95)ζ - 4.1 / n2 On the other hand, since the 5d1 consists of one electron, we apply the Bohr formula as E(5d1) = (-13.6057)ζ2/n2 Therefore E2 = 20.5 eV = -E(5d2) + E(5d1) = - (16.95)ζ + 4.1) / n2 Then using n = 6 the above equation can be written as (13.6057)ζ2 - (16.95)ζ - 733.9 = 0 and solving for ζ we get ζ = 8 < 11 . ' EXPLANATION OF E3 = 30 eV = -E(5d2) + E(5d1)' Here the E(5d2) represents the binding energy of the second 5d2, while the E(5d1) represents the binding energy of 5d1, which appears after the first ionization of the second 5d2 . Here the charges (-68e) of the following electron configuration (1s22s22p63s23p63d104s24p64d105s25p64f14) screen the nuclear charge (+79e ) and for a perfect screening we would have the same effective ζ = 11. Note that after the ionization of the first 5d2 the new electron cloud of 5d8 penetrates also strongly the 4f14 and provides an effective ζ < 11. As in the case of E2 here the second 5d2 consists of one pair (2 electrons of opposite spin). Thus applying my formula of 2008 we write -E(5d2) = -+ (16.95)ζ - 4.1 / n2 On the other hand, since the 5d1 consists of one electron, we apply the Bohr formula as E(5d1) = (-13.6057)ζ2/n2 Therefore E3 = 30 eV = -E(5d2) + E(5d1) = - (16.95)ζ + 4.1) / n2 Then using n = 6 the above equation can be written as (13.6057)ζ2 - (16.95)ζ - 1075.9 = 0 and solving for ζ we get ζ = 9.54 < 11. Of course the two electrons of opposite spin (5d2) do not provide any mutual repulsion, because I discovered in 2008 that at very short inter-electron separations the magnetic attraction is stronger than the electric repulsion giving a vibration energy, which seems to be like a simple electric repulsion of the Coulomb law. This situation of a vibration energy due to an electromagnetic interaction indeed occurs, because the peripheral velocity of a spinning electron is faster than the speed of light, which invalidates Einstein’s theory of special relativity. (See my FASTER THAN LIGHT). However under the influence of invalid relativity and in the absence of a detailed knowledge about the mutual electromagnetic interaction between the electrons of opposite spin today many physicists believe incorrectly that two electrons with opposite spin exert a mutual Coulomb repulsion. Under such fallacious ideas I published my paper of 2008 . Category:Fundamental physics concepts